Volumes of Revolution | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

A rotating strip makes a thin circular disc

Rotate a filled region through $360^\circ$ about the x-axis. A vertical strip sweeps out a thin cylinder, called a disc. Its radius is the distance from the axis to the curve, and its thickness is the small change in $x$ .

The ellipses show circular cross-sections viewed at an angle. Each disc has area πy² and small thickness along x.

A circle has area $\pi r^2$ . Multiplying by thickness gives a thin disc's volume. Adding thinner and thinner discs gives the exact solid volume:

V=\pi\int_a^b y^2\,dx

This formula assumes the region extends from the axis to the curve, with $a<b$ . The radius is $|y|$ , so its square is $y^2$ even below the axis. The formula is for volume, not surface area.

Square the radius before integrating

Worked example

Rotate the region under y = √x from x = 0 to x = 4 about the x-axis.

Here $y^2=x$ . Square first, then integrate:

\begin{gathered}V=\pi\int_0^4x\,dx\\=\pi\left[\frac{x^2}{2}\right]_0^4\\=8\pi\end{gathered}

The answer is in cubic units. It is smaller than the cylinder with radius 2 and length 4, whose volume is $16\pi$ . Keep $\pi$ in an exact answer.

Common mistake

Square the whole function, not each term separately or the final integral. If

y=x+1

, then

y^2=(x+1)^2=x^2+2x+1

Your turnPractice

The region between $y=x+1$ , the x-axis, $x=0$ and $x=2$ is rotated about the x-axis. Find the exact volume.

Show worked answer

\begin{gathered}\int_0^2(x^2+2x+1)\,dx\\=\left[\frac{x^3}{3}+x^2+x\right]_0^2=\frac{26}{3}\\V=\pi\times\frac{26}{3}=\frac{26\pi}{3}\end{gathered}